5  Group Homomorphisms Revisited

We now come to the central topic of Algebra. Homomorphisms have been at work behind the scenes in much of our work in this module as well as in the Abstract Algebra module. Informally, a homomorphism is a structure-preserving mapping from one group into another group, \(G\) into \(\gp\) say. Such mappings are used to provide information about the structure of \(\gp\) from the known structural properties of \(G\) and vice versa. The formal definition is as follows:

Definition 5.1 (Homomorphism) Let \(G\) and \(\gp\) be groups. A mapping \(\phi \, : \, G \rightarrow \gp\) is a homomorphism if, for all \(x, y \in G\) \[\phi (xy) = \phi(x) \phi(y).\]

Note

An isomorphism is simply a bijective homomorphism.

An easy example of structure preservation is the following lemma.

Lemma 5.1 Let \(\phi\) be a homomorphism from a group \(G\) into a group \(\gp\). Then,

1. if \(e\) is the identity in \(G\), then \(\phi (e)\) is the identity in \(\gp\),
2. for all \(x \in G\), \(\phi (x^{-1}) = (\phi (x))^{-1}\).

Proof.

We take each in turn.

1. Notice that \[\phi(x) = \phi(xe) = \phi(x)\phi(e).\] Pre-multiplying by \(\phi(x)^{-1}\), and writing \(e'\) for the identity of \(G'\) we have: \[ e' = (\phi(x)^{-1} \phi(x)) = \phi(x)^{-1}(\phi(x)\phi(e)) = \phi(e)\]as required.
2. We have: \[ e' = \phi(e) = \phi(xx^{-1}) = \phi(x)\phi(x^{-1}).\] Therefore \(\phi(x)^{-1} = \phi(x)\) as required.

Example 5.1  

The map \(\phi: \Z \to \Z_2\) by \(\phi(n) = \begin{cases} 0, & n \text{ is even} \\ 1, & n \text{ is odd} \end{cases}\). Then \(\phi\) is a homomorphism. This follows since the sum of two even or two odd numbers is even and the sum of an odd and even number is odd.

Example 5.2  

Two homomorphisms arise naturally in linear algebra:

1. Let \(V\) and \(W\) be vector spaces and \(T:V \to W\) be a linear transformation. Recall that \(V\) and \(W\) form abelian groups under vector addition \(+\). Thus \(T: (V,+) \to (W,+)\) is a homomorphism since \[T(u + v) = T(u) + T(v)\] for all \(u,v \in V\).
2. The set \(\glr{n}\) of all invertible \(n\times n\) matrices over the real numbers is a group under matrix multiplication. The map \(\det: \glr{n} \to (\R\backslash\{0\},\times)\) by \(A \mapsto \det(A)\) is a homomorphism since \(\det(AB) = \det(A)\det(B)\).

Example 5.3  

Similarly to Example 5.1 we can define a homomorphism \(\phi\) from the symmetric group \(S_n\) to \(\Z_2\) by: \[\phi(\sigma) = \begin{cases} 0, & \sigma \text{ is even }\\ 1, & \sigma \text{ is odd. } \end{cases}\]

Example 5.4  

The set \(\mathrm{C}[a,b]\) of all continuous functions on the interval \([a,b]\) forms a group under pointwise addition: \[f+g(x) = f(x) + g(x).\] The map \(\phi: \mathrm{C}[a,b] \to \R\), defined by \[\phi(f) = \int_{a}^{b} f(x) \ \dx\] is a group homomorphism for \[ \int_{a}^{b} (f+g)(x) \ \dx= \int_{a}^{b} f(x) + g(x) \ \dx = \int_{a}^{b} f(x) \ \dx + \int_{a}^{b} g(x) \ \dx.\]

Example 5.5  

Consider the group \(D_4\) and the normal subgroup \(N= \{e,r_1, r_2, r_3\}\). Define \(\phi: D_4 \to D_4/N\) by \(\phi(g) = gN\). Then \(\phi\) is a homomorphism from \(D_4\) to \(D_4/N\), since \[\phi(hg) = (hg)N = (hN)(gN) = \phi(h)\phi(g).\]

Notice that all elements of \(N\) (in \(D_4\)) map to the identity element \(N\) in the group \(D_4/N\).

Note that if we chose a non-normal subgroup \(H\) of \(D_4\) (e.g \(H= \{e,s_1\}\)), the map \(\phi\) from \(D_4\) to the set of left cosets of \(H\) given by \(\phi(g) = gH\) is a well-defined map but is not a homomorphism since the set of left cosets of \(H\) do not form a group (coset multiplication is only well-defined over cosets of a normal subgroup).

We shall see that there is an intimate connection between homomorphisms and normal subgroups. Before continuing, though, we make the following definition about sets.

Definition 5.2 (Image and Pre-image) Let \(f\) be a mapping of a set \(X\) into a set \(Y\), and let \(A \subseteq X\) and \(B \subseteq Y\). The image of \(A\) in \(Y\) under \(f\) is denoted and defined by \[f(A) = \{ f(a) \, | \, a \in A \}.\] The pre-image of \(B\) in \(Y\) under \(f\) is denoted and defined by \[f^{-1} (B) = \{ x \in X \, | \, f(x) \in B \}.\]

Our first observation is that the image of a group homomorphism is a group.

Theorem 5.1 Let \(\phi\) be a group homomorphism from a group \(G\) into a group \(\gp\). Then, for any subgroup \(H\) of \(G\), \(\phi (H)\) is a subgroup of \(\gp\).

Proof.

Since \(\phi(e)\) is the identity element \(e'\) of \(\gp\), then \(\phi(H)\) is non-empty.

Let \(g',h' \in \phi(H)\). Then there are elements \(g,h \in H\) such that \(\phi(g) = g'\) and \(\phi(h) = h'\). Thus: \[g'h' =\phi(g)\phi(h) = \phi(gh) \in \phi(H)\] and \(\phi(H)\) is closed under products.

Moreover, \[(g')^{-1} = (\phi(g))^{-1} = \phi(g^{-1}) \in \phi(H)\] and \(\phi(H)\) is closed under inverses.

5.1 Kernels and the First Isomorphism Theorem

In this section we establish the fundamental connection between homomorphisms and normal subgroups. We begin with a definition.

Definition 5.3 (Kernel) Let \(\phi\) be a group homomorphism from a group \(G\) into a group \(\gp\). The kernel of \(\phi\) is denoted and defined by \[ker(\phi) = \phi ^{-1}(e^{\prime}) = \{ x \in G \, | \, \phi(x) = e^{\prime} \},\] where \(e^{\prime}\) is the identity in \(\gp\).

The kernel of a homomorphism consists, therefore, of all of the elements of \(G\) that are mapped onto the identity in \(\gp\).

Lemma 5.2 Let \(\phi\) be a group homomorphism from a group \(G\) into a group \(\gp\). Then \(ker(\phi)\) is a normal subgroup of \(G\).

Proof.

First we need to demonstrate that \(ker(\phi)\) is a subgroup. Clearly, the \(ker(\phi)\) is non-empty since it contains the identity of \(G\).

Let \(g,h \in ker(\phi)\). Then \[\phi(gh) = \phi(g)\phi(h) = e' e' = e'\] for \(e'\) the identity element of \(\gp\). Therefore, \(ker(\phi)\) is closed under products. Moreover, \[\phi(g^{-1}) = \phi(g)^{-1} = (e')^{-1} = e'\] and \(ker(\phi)\) is closed under inverses.

These completes the subgroup tests. Now we show that \(ker(\phi)\) is a normal subgroup of \(G\).

Let \(h \in ker(\phi)\) and \(g \in G\). Then \[\begin{align*} \phi(gh\ginv) &= \phi(g)\phi(h)\phi(\ginv)\\ &= \phi(g)e'\phi(\ginv)\\ &= \phi(g)\phi(\ginv)\\ &= \phi(g\ginv)\\ &= \phi(e)\\ &=e'. \end{align*}\] Therefore, for all \(g \in G\), \(g \ker(\phi)\ginv = \ker(\phi)\) and so \(\ker(\phi)\) is a normal subgroup of \(G\).

Example 5.6  

The kernel of the map \(\phi: \Z \to \Z_2\) is precisely the even integers. (This is clearly a normal subgroup of \(\Z\) since \(\Z\) is abelian.)

Example 5.7  

Returning to Example 5.2:

1. The kernel of a linear transformation \(T: V \to W\) was defined in the linear algebra module as \[\krn(T) = \{v \in V : T(v) = 0\}.\] This definition coincides with the definition of the kernel of the group homomorphism \(T: (V,+) \to (W,+)\).

2. The kernel of the homomorphism \(\det: \glr{n} \to (\R\backslash\{0\},\times)\) is the set of all \(n\times n\) matrices with determinant \(1\).

Example 5.8  

The kernel of the homomorphism \(\phi: S_n \to \Z_2\) defined by \[\phi(\sigma) = \begin{cases} 0, & \sigma \text{ is even}\\ 1, & \sigma \text{ is odd} \end{cases}\] is precisely the normal subgroup \(A_n\) of \(S_n\).

Example 5.9  

The kernel of the homomorphism \(\phi: \mathrm{C}[a,b] \to \R\) of Example 5.4 is the set of elements of \(\mathrm{C}[a,b]\) whose integral over the interval \([a,b]\) is \(0\). (Note the kernel consists of not just the zero function.)

Example 5.10  

Revisiting Example 5.5, the kernel of the homomorphism \(\phi\) is precisely the normal subgroup \(N\).

Since \(ker(\phi)\) is a normal subgroup of \(G\), it follows that we can form the quotient group of \(G\) by \(ker(\phi)\). The following result tells us that this quotient group is isomorphic to the image of \(G\) under the homomorphism; this is known as the First Isomorphism Theorem (there are two more but we will not need them).

Theorem 5.2 (First Isomorphism Theorem) Let \(\phi : G \rightarrow \gp\) be a group homomorphism with kernel \(K\). Then \[\phi(G) \cong G/K.\]

Proof.

Define a map \(\psi: \phi(G) \to G/K\) by \(\phi(g) \mapsto gK\) for all \(g \in G\). We need to prove that: + that \(\psi\) is a well-defined map; + that \(\psi\) is a bijection; + that \(\psi\) is a homomorphism.

First we prove that \(\psi\) is well-defined. Let \(g,h \in G\) such that \(\phi(g) = \phi(h)\). We need to show that \(\psi(\phi(g)) = \psi(\phi(h))\). Since \(\phi(g)= \phi(h)\) then \(\phi(g h^{-1}) = \phi(g)\phi(h)^{-1} = e'\), where \('e\) is the identity element of \(\gp\). Therefore, \(gh^{-1} \in K\). Let \(k \in K\) be such that \(gh^{-1} = k\). Then, by postmultiplying on both sides by \(h\), we have \(g = kh\). We now have, \[\phi(g) = \phi(kh) = \phi(k)\phi(h) = e' \phi(h) = \phi(h)\] and \(\phi(g) = \phi(h)\) as required.

Clearly the map \(\psi\) is surjective since for any \(gK \in G/K\) \(\psi(\phi(g)) = gK\). Thus it remains only to show that \(\psi\) is a injective to conclude that it is bijective.

Suppose \(\phi(g), \phi(h) \in \phi(G)\) satisfy \(gK = hK\). We want to conclude that \(\phi(g) = \phi(h)\). Since \(gK = hK\), then \(gh^{-1} \in K\). Therefore there is a \(k \in K\) such that \(g = hk\) (repeating an argument above). Repeating the argument for well-definedness, we conclude that \(\phi(g)= \phi(h)\) as required.

The last is the most straightforward. We verify the homomorphism criteria: \[\psi(\phi(g)\phi(h)) = \psi(\phi(gh)) = ghK = gK hK = \psi(\phi(g))\psi(\phi(h))\] as required.

The converse of the First Isomorphism Theorem is also true, namely if \(N\) is a normal subgroup of \(G\) then we can easily define a group homomorphism \(\phi : G \rightarrow \gp\) such that \(G/N \cong \phi (G)\). The exact nature of the group \(\gp\) is not important and there are infinitely many groups \(\gp\) that we could choose to serve our purpose (any group containing a subgroup isomorphic to \(G/N\) will do), but the easiest one to choose is \(\gp = G/N\).

Lemma 5.3 Let \(N\) be a normal subgroup of \(G\). Then \(\phi : G \rightarrow G/N\) defined by \(\phi(x) = xN\), for all \(x \in G\), is a homomorphism with kernel \(N\).

Proof.

The map \(\phi\) is a homomorphism since \[\phi(xy) = (xy)N = xN yN = \phi(x)\phi(y).\]

Clear \(N \subseteq \ker(\phi)\). On the other hand, if \(x \in \ker(\phi)\) then \(xN = N\) and so \(x \in N\). Thus \(\ker(\phi) = N\) as required.

The map \(\phi\) in the statement of Lemma 5.3 is called the natural or canonical homomorphism.

In summary, the image of a homomorphism of \(G\) is isomorphic to \(G/N\) where \(N\) is the kernel of the homomorphism, and for every normal subgroup \(N\) of \(G\) there is a homomorphism from \(G\) to \(G/N\) with kernel \(N\) (each homomorphism has a different codomain as \(N\) changes). Group homomorphisms and quotient groups are really just two different ways of looking at the same thing. Compare the following example and theorem to those in Chapter 4.

Example 5.11 Classify the group \((\Z_4 \times \Z_2) / \langle (0, 1) \rangle\) according to the Fundamental Theorem of Finite Abelian Groups.

The map \(\phi:\Z_4 \times \Z_2 \to \Z_4\) defined by \(\phi((x,y)) = x\) is a homomorphism with kernel \[\{0\} \times \Z_2 = \{(0,0),(0,1)\}.\] Therefore, using the First Isomorphism Theorem, \((\Z_4 \times \Z_2) / \langle (0, 1) \rangle \cong \Z_4\).

In very loose terms the previous example can be represented as follows: \[(\Z_4 \times \Z_2) / \langle (0, 1) \rangle = \frac{\Z_4 \times \Z_2}{\{0\} \times \Z_2} = \Z_4.\]

Theorem 5.3 Let \(G = H \times K\) be the direct product of groups \(H\) and \(K\). Then \(\overline{K} = \{ (e, k) \, | \, k \in K \}\) is a normal subgroup of \(G\) isomorphic to \(K\) and \(G/\overline{K} \cong H\).

Proof.

The map \(\phi: H \times K \to H\) by \((h,k) \mapsto h\) is easily verified to be a group homomorphism. The kernel of \(\phi\) is precisely the set \[\{e_{H}\} \times K = \{(e_H, k) : K \in K\}\] where \(e_{H}\) is the identity element of \(H\).

By the First Isomorphism Theorem, we have \(\phi(H \times K) = H \cong G/\overline{K}\).

5.2 Homomorphisms and Group Actions

When introducing group actions we made the following remark:

Groups acting on sets are really just thinly disguised permutation groups in that whenever a group \(G\) acts on a set \(X\) we can associate with each \(g\in G\) the permutation \(f_g\) of \(X\) defined by \(f_g(x)=g*x,\ \forall \, x\in X\). These permutations form a subgroup of \(S_{|X|}\), but this group need not be isomorphic to \(G\), and we shall return to this later.

We are now in a position to state precisely what was meant by this, and we begin with the following theorem (note that, as usual, we denote the action of a group element \(g\) on an element \(x \in X\) by \(g * x\)):

Theorem 5.4 Let \(G\) be a group acting on a finite set \(X\). Define \(\phi : G \rightarrow S_{|X|}\) by \(\phi(g) = f_g\) where \(f_g(x) = g*x\) for all \(x \in X\). Then \(\phi\) is a homomorphism.

Proof.

We need to show that the map \(\phi\) is well-defined. In this context, this means verifying that \(f_g\) is in fact a bijection on \(X\). A map is bijective if and only if it is invertible and so it suffices to show that \(f_g\) is invertible. We have: \[(f_g \circ f_{g^{-1}})(x) = f_g(f_{g^{-1}}(x)) = f_g(g^{-1}\ast x) = g \ast(g^{-1}\ast x) = e\ast x = x.\] Similarly one has that \(f_{\ginv} \circ f_{g}(x) =x\) for all \(x \in X\). Therefore \(f_g\) is invertible (with inverse \(f_{\ginv}\)) and so it is a bijective map on \(X\).

It now remains to verify that \(\phi\) is a homomorphism. We have \[\phi(gh)(x) = f_{gh}(x) = (gh)\ast x = g\ast(h \ast x) =f_g(f_h(x)) = \phi(g)(\phi(h)(x))\] for all \(x \in X\). This means that \(\phi(gh)= \phi(g)\phi(h)\) and so \(\phi\) is a homomorphism.

Remark. The kernel of the homomorphism in Theorem 5.4 is precisely the set of elements \(g \in G\) with \(\fix(g) = X\).

Example 5.12  

Recall that in Chapter 3, Example 3.5 we considered the group \(D_4\) acting on the diagonals \(d_1\) and \(d_2\) of the square.

We demonstrated that the permutation group corresponding to this action is: \[ \left\{ \begin{pmatrix} d_1 & d_2 \\ d_1 & d_2 \end{pmatrix},\begin{pmatrix} d_1 & d_2 \\ d_2 & d_1 \end{pmatrix} \right\} \cong S_2. \]

By Theorem 5.4, \(\phi: D_4 \to S_2\) is defined by: \[\begin{align*} \phi(e)&= \phi(r_2)=\phi(s_2) = \phi(s_4) = \begin{pmatrix} 1 & 2 \\ 1 & 2 \end{pmatrix} \\ \phi(r_1)&= \phi(r_3)=\phi(s_1) = \phi(s_3) = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} \end{align*}\] is a homomorphism with kernel \(K = \{ e,r_2, s_2, s_4\}\). It follows that \(K\) is a normal subgroup of \(D_4\) and \(D_4/K \cong S_2\).

5.3 Problem Sheet 5

For Week 11; covers Chapter 5.

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Question 5.1

Let \(\phi : \Z_4 \times \Z_6 \rightarrow D_4\) be a homomorphism with kernel \(K = \langle (2, 2) \rangle\), where \(D_4\) denotes the dihedral group of order 8.

1. List the elements of \(K\) and of \((\Z_4 \times \Z_6) / K\).
2. Classify \(\phi (\Z_4 \times \Z_6)\) according to the Fundamental Theorem of Finite Abelian Groups.

Solution 5.1

1. \(K = \{(0, 0), (0, 2), (0, 4), (2, 0), (2, 2), (2,4)\}\). \((\Z_4 \times \Z_6) / K = \{ K, (1, 0) + K, (0, 1) + K, (1, 1) + K\}\).
2. Since \(| \, (\Z_4 \times \Z_6) / K \, |\) is 4, then this group is isomorphic to either \(\Z_4\) or \(\Z_2 \times \Z_2\). Note that each non-identity element of \((\Z_4 \times \Z_6) / K\) has order 2 (convince yourselves that this is the case) and so \((\Z_4 \times \Z_6) / K \cong \Z_2 \times \Z_2\). Hence \(\phi (\Z_4 \times \Z_6) \cong \Z_2 \times \Z_2\).

Question 5.2

Let \(\phi : G \rightarrow \gp\) be a group homomorphism. Show that if \(|G|\) and \(|\gp|\) are finite, then:

1. \(| \phi (G) |\) is a divisor of \(|\gp|\).
2. \(| \phi (G) |\) is a divisor of \(|G|\).
3. \(\forall \, g \in G\), the order of \(\phi (g)\) divides the order of \(g\).

Solution 5.2

1. \(\phi(G)\) is a subgroup of \(\gp\) and so, by Lagrange’s Theorem, \(| \phi (G) |\) divides \(|\gp|\).
2. We know that \(\phi(G) \cong G/K\), where \(K\) is the kernel of \(\phi\). As these two groups are isomorphic it follows that they must have the same order and so \[| \, \phi(G) \, | = | \, G/K \, | = \frac{| \, G \, |}{| \, K \, |}\] and hence \(| \, K \, | = \D \frac{| \, G \, |}{| \, \phi(G) \, |}\). It then follows that \(| \, \phi(G) \, |\) divides \(| \, G \, |\).
3. We have that \(\phi(g^m) = (\phi(g))^m\) for all \(m \in \Z\). So, if \(g\) has order \(k\) in \(G\), then \[e^{\prime} = \phi(e) = \phi(g^k) = (\phi(g))^k\] and so the order of \(\phi(g)\) divides \(k\) (recall that the order of an element is the smallest positive integer power, \(n\) say, such that \(\phi(g))^n = e^{\prime}\). So, if \((\phi(g))^k = e^{\prime}\), then \(n\) must divide \(k\)).

Question 5.3

Let \(\phi : G \rightarrow \gp\) be a group homomorphism. For each of the following statements, show whether it is true or false by providing either a proof or a non-trivial counter-example as appropriate. (The trivial homomorphism from \(G\) to \(\gp\) is defined by \(\phi (g) = e^{\prime}, \; \forall \, g \in G\).)

1. If \(G\) is abelian, then \(\gp\) is abelian.
2. If \(\gp\) is abelian, then \(G\) is abelian.
3. If \(G\) is abelian, then \(\phi (G)\) is abelian.
4. If \(\phi (G)\) is abelian, then \(G\) is abelian.

Solution 5.3

1. This statement is . Consider the mapping \(\phi : \Z_2 \rightarrow \Z_2 \times D_3\) defined by \(\phi(0) = (0, e)\) and \(\phi(1) = (1, e)\). This is a homomorphism since \[\phi(0 + 1) = \phi(1) = (1, e) = (0, e)(1, e) = \phi(0) \phi(1),\] and whilst \(\Z_2\) is abelian, \(\Z_2 \times D_3\) is not abelian.
2. This statement is . Consider the mapping \(\phi : D_3 \rightarrow \Z_2\) defined by \(\phi(e) = \phi(r_1) = \phi(r_2) = 0\) and \(\phi(s_i) = 1\) for all \(i = 1, 2, 3\). Clearly this is a homomorphism (convince yourselves) and whilst \(\Z_2\) is abelian, \(D_3\) is non-abelian.
3. This statement is . Let \(\phi(x), \phi(y) \in \phi(G)\). Then, \[\begin{align*} \phi(x)\phi(y) &= \phi(xy) \qquad \qquad \text{(homomorphism)} \\ &= \phi(yx) \qquad \qquad \text{($G$ is abelian)} \\ &= \phi(y)\phi(x), \end{align*}\] so \(\phi(G)\) is abelian.
4. This statement is . The counter example in part (b) above will suffice. (If you want a non-surjective homomorphism then you could have \(\phi : D_3 \rightarrow \Z_2 \times \Z_2\) defined by \(\phi(e) = \phi(r_1) = \phi(r_2) = (0,0)\) and \(\phi(s_i) = (1,0)\) for all \(i = 1, 2, 3\).)

Question 5.4

By considering all of the normal subgroups of \(D_4\), determine the number of distinct homomorphisms that exist from \(D_4\) to \(A_4\).

Solution 5.4

The first thing to note is that as the kernel of a homomorphism mapping from \(G\) to \(\gp\) is a normal subgroup of \(G\) and, therefore, in this example, must be one of the normal subgroups of \(D_4\). There are six normal subgroups of \(D_4\) which are \(D_4\), \(\{e\}\), \(\{e, r_2\}\), \(\{e, r_1, r_2, r_3\}\), \(\{e, r_2, s_1, s_3\}\), and \(\{e, r_2, s_2, s_4\}\). If \(\phi : D_4 \rightarrow A_4\) is a homomorphism, then \(K = \text{ker}\, (\phi)\) is a normal subgroup of \(D_4\) and \(\phi(D_4)\) is a subgroup of \(A_4\) (Theorem 5.1), with \[|K| \times | \phi(D_4)| = 8.\] (This follows from the First Isomorphism Theorem; as \(\phi(G) \cong G/K\) then the sizes of these two groups must be the same, so \(|K| \times |\phi(G)| = |G|\).) Clearly, \(K \neq \{e\}\) since \(A_4\) has no subgroup of order 8 (\(|A_4| = 12\)). If \(K=D_4\), his implies that \(\phi\) is the trivial homomorphism. If \(|K|=4\) then \(|\phi(D_4)|=2\), so \(\phi(D_4) = \{ \,\text{id}, (1 \; 2)(3\; 4 )\}\), \(\{ \,\text{id}, (1 \; 3)(2\; 4 )\}\), or \(\{ \,\text{id}, (1 \; 4)(2\; 3 )\}\), giving three homomorphisms for each of the three possible choices of \(K\) (there are three normal subgroups of \(D_4\) of order 4). Hence, there are 9 distinct homomorphisms of this type. If \(|K|=2\) then \(K=\{e, r_2\}\) and \[D_4/K = \{ \{e, r_2\}, \{r_1, r_3\}, \{s_1, s_3\}, \{s_2, s_4\}\},\] which has order 4. So \(D_4/K \cong \Z_2 \times \Z_2\) as each non-identity element in \(D_4/K\) has order 2. Hence, \(\phi(D_4)=\{ \,\text{id}, (1 \; 2)(3\; 4 ), (1 \; 3)(2 \; 4), (1 \; 4)(2 \; 3)\}\) by the First Isomorphism Theorem. We know that \(\phi(e) = \phi(r_2) = \, \text{id}\), \(\phi(r_1) = \phi(r_3)\), \(\phi(s_1) = \phi(s_3)\), and \(\phi(s_2) = \phi(s_4)\). There are no further restrictions, so there are 6 homomorphisms of this type. Hence the total number of homomorphisms is \(1+9+6 = 16\).

Question 5.5

Let \(G,H,K\) be groups.

1. If \(\phi:G \to H\) is a homomorphism and \(\psi: H \to K\) is a homomorphism, show that \(\psi\phi: G \to K\).
2. If \(\phi: G \to H\) is an isomorphism, show that \(\phi^{-1}: H \to G\) is also an isomorphism.

Solution 5.5

1. Let \(x,y \in G\). Then \[\psi\phi(xy) = \psi(\phi(xy))= \psi(\phi(x)\phi(y)) = \psi(\phi(x))\psi(\phi(y)) = \psi\phi(x)\psi\phi(y)\] as required.
2. Clearly \(\phi^{-1}\) is also a bijection. Therefore it remains to show that \(\phi^{-1}\) is a homomorphism. Let \(s,t \in H\). There are \(x,y \in G\) such that \(\phi(x)= s\) and \(\phi(y) = t\). We have \[\phi^{-1}(st) = \phi^{-1}(\phi(x)\phi(y)) = \phi^{-1}(\phi(xy)) = xy = \phi^{-1}(s)\phi^{-1}(t)\] as required.

Question 5.6

Let \(G\) be a group and fix an element \(g \in G\). Define a mapping \(\kappa_g: G \to G\) by \(\kappa_g(h) = ghg^{-1}\) for all \(h \in G\). Show that \(\kappa_g\) is an isomorphism from \(G\) to \(G\).

Solution 5.6

First we show that \(\kappa_g\) is invertible (and so a bijection). Define \(\kappa_{g^{-1}}: G \to G\) by \(\kappa_{g^{-1}}(h) = g^{-1}hg\) for all \(h \in G\). Observe that for any \(h \in G\),

\[\kappa_{g}\kappa_{g^{-1}}(h) = \kappa_g(g^{-1}hg) = g(g^{-1}hg)g^{-1} = h.\] Similarly, \(\kappa_{g^{-1}}\kappa_{g}(h)=h\) for all \(h \in G\).

It follows that \(\kappa_{g^{-1}} = \kappa_{g}^{-1}\). Therefore, \(\kappa_{g}\) is invertible and so a bijection.

Now we show that \(\kappa_{g}\) is a homomorphism.

Let \(h,k \in G\), then, \[\kappa_{g}(hk) = ghkg^{-1} = gh(g^{-1}g)kg^{-1} = (ghg^{-1})(gkg^{-1}) = \kappa_{g}(h)\kappa_{g}(k)\] as required.

Question 5.7

Write down a non-trivial homomorphism from \(A_4\) to \(D_3\) and describe its kernel, \(K\). Form the quotient group \(A_4/K\) and show directly that this is isomorphic to the image of \(A_4\). (This is a bit tricky; remember that the image of \(A_4\) need not be the whole of \(D_3\), but must be a non-trivial subgroup of \(D_3\).)

Solution 5.7

If \(\phi : A_4 \rightarrow D_3\) is a homomorphism, then \(\phi(A_4)\) is a subgroup of \(D_3\) isomorphic to \(A_4/K\) (First Isomorphism Theorem). Now, \(|A_4| = 12\) so, by Lagrange’s Theorem, we know that \(|K| = 1, 2, 3, 4, 6\) or \(12\). \(|K| = 1 \imp | \, A_4/K \, | = 12\) and \(|K| = 3 \imp | \, A_4/K \, | = 4\), but neither of these are possible since \(D_3\) does not have subgroups of orders 12 or 4 (as \(|D_3| = 6\)). \(|K| = 12 \imp | \, A_4/K \, | = 1\), so \(\phi\) is trivial. \(|K| \neq 6\) since \(A_4\) has no subgroup of order 6. Now, \(A_4\) has eight elements of order 3 and three of order 2 (cyclic structures \((* \; * \; *)\) and \((* \; *)(* \; *)\) respectively - see Problem Sheet 3 Q4). By part 1(c) above, the elements of order 3 must map to the identity or one of the rotations in \(D_3\) and, similarly, the elements of order 2 must map onto the identity or a reflection. Now, if \(|K| = 2\) then \(A_4/K \cong D_3\) but this is not possible since \(A_4 / K\) only has two elements of order 2 whereas \(D_3\) has three elements (each of the reflections) of order 2. We now consider why this is so. Any subgroup of order 2 in \(A_4\) must contain the identity and one element of order 2. So, there are only three possible subgroups of \(A_4\) of order 2, namely \(\{id, (1 \; 2)(3 \; 4) \}, \{id, (1 \; 3)(2 \; 4) \}\) and \(\{id, (1 \; 4)(2 \; 3) \}\). Now, a coset, \(\sigma K\) from \(A_4/K\) has order 2 if and only if \(\sigma \notin K\) (if it were then \(\sigma K = K\) and that has order 1 in \(A_4/k\)), and \(\sigma ^2 \in K\). We illustrate this last point by example. Suppose that \(K = \{ id, (1 \; 2)(3 \; 4)\}\), how can we form a \(\sigma K\) of order 2 in \(A_4/K\)? Well, \((1 \; 3)(2 \; 4)K = (1 \; 4)(2 \; 3)K\) and this coset has period 2, but there are no others. Also, any other coset \(\sigma K\) such that \(\sigma = (* \; * \; *)\) does not have order 2 since \(\sigma ^2 \neq \text{id}\) nor is \(\sigma ^2 \in K\). So, there is exactly one element of period 2 in \(A_4/K\), and so \(A_4/K \ncong D_3\). If \(|K| = 4\) then \(|A_4/K| = 3\) and hence \(A_4/K \cong \{e, r_1, r_2 \}\). Now, \(K = \{ e, (1 \; 2)(3 \; 4), (1 \; 3)(2 \; 4), (1 \; 4)(2 \; 3)\}\) and \(A_4/K = \{K, (1 \; 2 \: 3)K, (1 \; 2 \; 4)K \}\). We now define a homomorphism \(\phi : A_4 \rightarrow D_3\) by: \(\phi(e) = \phi((1 \; 2)(3 \; 4)) = \phi((1 \; 3)(2 \; 4)) = \phi((1 \; 4)(2 \; 3)) = e\) \(\phi(\sigma) = r_1 \quad \forall \, \sigma \in (1 \; 2 \; 3)K = \{(1 \; 2 \; 3), (1 \; 3 \; 4), (2 \; 4 \; 3), (1 \; 4 \; 2) \}\) \(\phi(\sigma) = r_2 \quad \forall \, \sigma \in (1 \; 2 \; 4)K = \{(1 \; 2 \; 4), (1 \; 4 \; 3), (1 \; 3 \; 2), (2 \; 3 \; 4) \}\) This then forces the isomorphism (see the proof to the First Isomorphism Theorem) \(\mu : A_4/K \rightarrow \phi (D_3)\) (where \(\phi (D_3) = \{e, r_1, r_2 \}\)) defined by \(\mu (aK) \rightarrow \phi(a)\).